#2**+2 **

I think you have misunderstood the question guest ://

j is not a complex number.

x^2-3x+2j=0

Solutions for x

\(x=\frac{3\pm\sqrt{9-8j}}{2}\)

The roots will be complex when the discriminant is negative.

That is

\(9-8j<0\\ 9<8j\\ 9/8 9/8\\ j>1.125\)

So the solutions (for x) of x^2-3x+2j=0 will be complex if j>1.125

Melody
Oct 3, 2017

#1**+1 **

I replaced your "j" with "i", which is a more commonly used "complex number"

Solve for x:

2 i - 3 x + x^2 = 0

Subtract 2 i from both sides:

x^2 - 3 x = -2 i

Add 9/4 to both sides:

x^2 - 3 x + 9/4 = 9/4 - 2 i

Write the left hand side as a square:

(x - 3/2)^2 = 9/4 - 2 i

Take the square root of both sides:

x - 3/2 = sqrt(9/4 - 2 i) or x - 3/2 = -sqrt(9/4 - 2 i)

Add 3/2 to both sides:

x = sqrt(9/4 - 2 i) + 3/2 or x - 3/2 = -sqrt(9/4 - 2 i)

Add 3/2 to both sides:

**x = 3/2 + 1/2 sqrt(9 - 8 i) or x = 3/2 - sqrt(9/4 - 2 i)**

Guest Oct 3, 2017

#2**+2 **

Best Answer

I think you have misunderstood the question guest ://

j is not a complex number.

x^2-3x+2j=0

Solutions for x

\(x=\frac{3\pm\sqrt{9-8j}}{2}\)

The roots will be complex when the discriminant is negative.

That is

\(9-8j<0\\ 9<8j\\ 9/8 9/8\\ j>1.125\)

So the solutions (for x) of x^2-3x+2j=0 will be complex if j>1.125

Melody
Oct 3, 2017